Juestion 1: he load is supported by the four 304 stainless stell wires that are connected to the rigid members AB and DC. Determine the angle of tilt of each member after the C (Newton) load is applied. he members were originally horizontal, and each wire has a cross-sectional area of 15 mm. E304=193 GPa E F Cake: A= 1.2 m meter B= 2.2 m C= 2497 N B meter 1 Fo.5 m 1 m 0.9 m B 1.5 m 0.5 m

Elements Of Electromagnetics
7th Edition
ISBN:9780190698614
Author:Sadiku, Matthew N. O.
Publisher:Sadiku, Matthew N. O.
ChapterMA: Math Assessment
Section: Chapter Questions
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Question 1:
The load is supported by the four 304 stainless stell wires that are connected to the rigid members
AB and DC. Determine the angle of tilt of each member after the C (Newton) load is applied.
The members were originally horizontal, and each wire has a cross-sectional area of 15 mm.
E304=193 GPa
E
F
Take:
A=
1.2
m
A meter
B=
2.2
m
B meter
C=
2497
N
10.5 m
1 m
0.9 m
0.5 mt
1.5 m
Solution:
1m
FDE
0.5m
0.5 m
1m
FBG
FAH
1.5m
0.5m
1.5m
0.5 m
C
Transcribed Image Text:Question 1: The load is supported by the four 304 stainless stell wires that are connected to the rigid members AB and DC. Determine the angle of tilt of each member after the C (Newton) load is applied. The members were originally horizontal, and each wire has a cross-sectional area of 15 mm. E304=193 GPa E F Take: A= 1.2 m A meter B= 2.2 m B meter C= 2497 N 10.5 m 1 m 0.9 m 0.5 mt 1.5 m Solution: 1m FDE 0.5m 0.5 m 1m FBG FAH 1.5m 0.5m 1.5m 0.5 m C
Internal Forces in the wires:
EM, = 0; 2F3G -C(1.5)= 0
EF, = 0; FAH +F30 -C = 0
F30
1877.25
N
F
N
AH
EM, = 0; 1.5FCF -0.5F = 0
EF, = 0; FDE +Fg - FAH = 0
F =
AH
FDE
N
Displacement :
FOELDE-
mm
ApgE
Feg LeF
AcgE
%3D
mm
mm
%3D
1.5
8g = 8g. + dc =
%3D
mm
tan a =
Ans: a =
1500
FLAE =
mm
AE
HF.
8, = 84 +84H =
mm
BG
mm
BG
tan B =
Ans: B=
2000
I|||
Transcribed Image Text:Internal Forces in the wires: EM, = 0; 2F3G -C(1.5)= 0 EF, = 0; FAH +F30 -C = 0 F30 1877.25 N F N AH EM, = 0; 1.5FCF -0.5F = 0 EF, = 0; FDE +Fg - FAH = 0 F = AH FDE N Displacement : FOELDE- mm ApgE Feg LeF AcgE %3D mm mm %3D 1.5 8g = 8g. + dc = %3D mm tan a = Ans: a = 1500 FLAE = mm AE HF. 8, = 84 +84H = mm BG mm BG tan B = Ans: B= 2000 I|||
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